Tensor is a right exact functor Elementary Proof

This is a relatively elementary proof (compared to others out there) of the fact that tensor is a right exact functor. Proof is taken from Hungerford, and reworded slightly. The key prerequisites needed are the universal property of quotient and of tensor product.

Statement

If $A\xrightarrow{f}B\xrightarrow{g}C\to 0$ is an exact sequence of left modules over a ring $R$ and $D$ is a right $R$ -module, then $\displaystyle D\otimes_R A\xrightarrow{1_D\otimes f}D\otimes_R B\xrightarrow{1_D\otimes g}D\otimes_R C\to 0$ is an exact sequence of abelian groups. An analogous statement holds for an exact sequence in the first variable.

Proof

(Hungerford 210)

We split our proof into 3 parts: (i) $\text{Im}(1_D\otimes g)=D\otimes_R C$ ; (ii) $\text{Im}(1_D\otimes f)\subseteq\text{Ker}(1_D\otimes g)$ ; and (iii) $\text{Ker}(1_D\otimes g)\subseteq\text{Im}(1_D\otimes f)$ .

(i) Since $g$ is an epimorphism by hypothesis every generator $d\otimes c$ of $D\otimes_R C$ is of the form $d\otimes g(b)=(1_D\otimes g)(d\otimes b)$ for some $b\in B$ . Thus $\text{Im}(1_D\otimes g)$ contains all generators of $D\otimes_R C$ , hence $\text{Im}(1_D\otimes g)=D\otimes_R C$ .

(ii) Since $\text{Ker} g=\text{Im} f$ we have $gf=0$ and $\displaystyle (1_D\otimes g)(1_D\otimes f)=1_D\otimes gf=1_D\otimes 0=0,$ hence $\text{Im}(1_D\otimes f)\subseteq\text{Ker}(1_D\otimes g)$ .

(iii) Let $\pi:D\otimes_R B\to(D\otimes_R B)/\text{Im}(1_D\otimes f)$ be the canonical epimorphism. From (ii), $\text{Im}(1_D\otimes f)\subseteq\text{Ker}(1_D\otimes g)$ so (by universal property of quotient Theorem 1.7) there is a homomorphism $\alpha:(D\otimes_R B)/\text{Im}(1_D\otimes f)\to D\otimes_R C$ such that $\displaystyle \alpha(\pi(d\otimes b))=(1_D\otimes g)(d\otimes b)=d\otimes g(b).$ We shall show that $\alpha$ is an isomorphism. Then $\text{Ker}(1_D\otimes g)=\text{Im}(1_D\otimes f)$ .

We show first that the map $\beta:D\times C\to(D\otimes_R B)/\text{Im}(1_D\otimes f)$ given by $(d,c)\mapsto\pi(d\otimes b)$ , where $g(b)=c$ , is independent of the choice of $b$ . Note that there is at least one such $b$ since $g$ is an epimorphism. If $g(b')=c$ , then $g(b-b')=0$ and $b-b'\in\text{Ker} g=\text{Im} f$ , hence $b-b'=f(a)$ for some $a\in A$ . Since $d\otimes f(a)\in\text{Im}(1_D\otimes f)$ and $\pi(d\otimes f(a))=0$ , we have
$\begin{aligned} \pi(d\otimes b)&=\pi(d\otimes(b'+f(a))\\ &=\pi(d\otimes b'+d\otimes f(a))\\ &=\pi(d\otimes b')+\pi(d\otimes f(a))\\ &=\pi(d\otimes b'). \end{aligned}$

Therefore $\beta$ is well-defined.

Verify that $\beta$ is middle linear:
$\begin{aligned} \beta(d_1+d_2,c)&=\pi((d_1+d_2)\otimes b)\qquad\text{where }g(b)=c\\ &=\pi(d_1\otimes b+d_2\otimes b)\\ &=\pi(d_1\otimes b)+\pi(d_2\otimes b)\\ &=\beta(d_1,c)+\beta(d_2,c). \end{aligned}$

$\begin{aligned} \beta(d,c_1+c_2)&=\pi(d\otimes(b_1+b_2))\qquad\text{where }g(b_i)=c_i\\ &=\pi(d\otimes b_1+d\otimes b_2)\\ &=\pi(d\otimes b_1)+\pi(d\otimes b_2)\\ &=\beta(d,c_1)+\beta(d,c_2). \end{aligned}$

Let $r\in R$ .
$\begin{aligned} \beta(dr,c)&=\pi(dr\otimes b)\qquad\text{where }g(b)=c\\ &=\pi(d\otimes rb)\\ &=\beta(d,rc)\qquad\text{where }g(rb)=rg(b)=rc. \end{aligned}$

By universal property of tensor product there exists a unique homomorphism $\bar{\beta}:D\otimes_R C\to(D\otimes_R B)/\text{Im}(1_D\otimes f)$ such that $\bar{\beta}(d\otimes c)=\beta(d,c)=\pi(d\otimes b)$ , where $g(b)=c$ .

Therefore, for any generator $d\otimes c$ of $D\otimes_R C$ , $\displaystyle \alpha\bar{\beta}(d\otimes c)=\alpha\pi(d\otimes b)=d\otimes g(b)=d\otimes c,$ hence $\alpha\bar{\beta}$ is the identity map.

Similarly
$\begin{aligned} \bar{\beta}\alpha(d\otimes b+\text{Im}(1_D\otimes f))&=\bar{\beta}\alpha\pi(d\otimes b)\\ &=\bar{\beta}(d\otimes g(b))\\ &=\pi(d\otimes b)\\ &=d\otimes b+\text{Im}(1_D\otimes F) \end{aligned}$
so $\bar{\beta}\alpha$ is the identity so that $\alpha$ is an isomorphism.